MICROPROCESSOR AND MICROCONTROLLER: Electric Field Intensity

Consider a fixed charge Q1. Let us move a second charge say test charge Qt around the Q1. We observe that the test charge Qt experiences force everywhere around Q1. It is experiencing a force field. Force on it will be given by coulombs law as:
$F_{t} = \frac{Q_{1}Q_{t}}{4\pi \epsilon _{0}R_{1t}^{2}}a_{1t}$
But still question arises what is Electric Field?? Before defining electric field let us modify above equation a bit.
Let us write the above equation as force/charge equation:
$\frac{F_{t}}{Q_{t}} = \frac{Q_{1}}{4\pi \epsilon _{0}R_{1t}}a_{1t}$
This quantity that is force/charge describes a vector quantity. This vector quantity is called Electric Field Intensity.So Electric field intesity is defined as the vector force on an unit positive test charge by a fixed charge Q1. Or it can also be defined as force per unit charge.
So to sum it up we can Electric field intensity as:
$E = \frac{F}{Q}$
So it becomes.
$E = \frac{Q}{4\pi \epsilon _{0}R^{2}}a_{R}$
Expressing Electric field in cartesian Coordinate:
$E(r) = \frac{Q}{4\pi \epsilon _{0}\left | r - r^{'} \right |^{2}}\frac{r - r^{'}}{\left | r - r^{'} \right |} = \frac{Q[(x - x^{'})i + (y - y^{'})j + (z - z^{'})k]}{4\pi \epsilon _{0}[(x - x^{'})^{2} + (y - y^{'})^{2} + (z - z^{'})^{2}]^{\frac{3}{2}}}$
Expressing Electric Field in Spderical Co-ordinates:
First for considering spherical co ordinate we must see that fixed charge has spherical symmetry at its location like at origin(0,0,0) for that we can either use spherical co-ordinate directly in equation as:
$E_{r} = \frac{Q_{1}}{4\pi \epsilon _{0}r^{2}}$
Where Er and r spherical co-ordinates values or we can convert them in cartesian coordinate which will give us the equation as below:
We will have r = R = xi + yj +zk ar = aR = (xi + yj + zk)/√(x^2+y^2+z^2 )
$E = \frac{Q}{4\pi \epsilon _{0}(x^{2}+y^{2}+z^{2})}(\frac{x}{\sqrt{x^{2}+y^{2}+z^{2}}} + \frac{y}{\sqrt{x^{2}+y^{2}+z^{2}}} + \frac{z}{\sqrt{x^{2}+y^{2}+z^{2}}})$
If more charges are added in the surrounding of Q1 then electric field will be sum of individual acting alone. As shown in equation given below:
$E(r) = \frac{Q_{1}}{4\pi \epsilon _{0}\left | r - r_{1} \right |^{2}}a_{1} + \frac{Q_{2}}{4\pi \epsilon _{0}\left | r - r_{2} \right |^{2}}a_{2} + .... + \frac{Q_{n}}{4\pi \epsilon _{0}\left | r - r_{n} \right |^{2}}a_{n}$

Before ending the topic let us do a small quick example.
Example: See the figure below. We need to find out Electric field at P(1,1,1) caused by 4 identical 3nC charges located at P1(1,1,0), P2(-1,1,0), P3(-1,-1,0) and P4(1,-1,0).

Example of Electric field

[caption id="" align="alignnone" width="615" caption="Example of Electric Field"][/caption]

Answer: As r = i + j + k and r1 = i+j therefore r – r1 = k. Therefore |r – r1| = 1.
Similaraly r – r2 = 2i + k. therefore |r – r2| = √5
Similaraly r – r3 = 2i + 2j + k. therefore |r – r3| = 3
Similaraly r – r4 = 2j + k. therefore |r – r4| = √5
And as
$\frac{Q}{4\pi 8.854*10^{-12}} = \frac{3*10^{-9}}{4\pi 8.854*10^{-12}} = 26.96 Vm$
$E = 26.96[\frac{k*1}{1*1^{2}}+\frac{2i + k}{\sqrt{5}}\frac{1}{\sqrt{5}^{2}} + \frac{2i + 2j + k}{3}\frac{1}{3^{2}}+ \frac{2j + k}{\sqrt{5}}\frac{1}{\sqrt{5}^{2}}]$

Therefore simplifying we get E =
E = 6.82i + 6.83j + 32.8k V/m ans.

MICROPROCESSOR AND MICROCONTROLLER

Saturday, August 13, 2011

Electric Field Intensity

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